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Twin paradox in Big Crunch

  1. Jul 28, 2009 #1
    I already posted about this under a different thread (Time dilation for clock thrown up and caught back) but, after reading some of the replies, figured that that was not the right way of attacking my real problem, which is figuring out what happens to time during a Big Crunch. So I will restate the actual problem:

    On the expanding universe, one might use two different metrics:
    1. A metric that doesn't contain any information about the expanding universe, and which pretty much obeys Special Relativity. Stuff just happens to be flying apart, but the metric does not treat this in a special way. Nothing travels faster than light. The universe is observed from our position, and time for distant objects is moving more slowly because they are moving relative to us. The speed of light is c everywhere.
    2. A metric that defines the time coordinate to be the local time experienced by a local observer who is moving together with the expansion of the universe. This is the "Cosmological Model", which cannot be experienced by any observer (except, some might say, God) but is very handy because the universe is homogenous, everything is the same age. The expansion does not cause any time dilation, by definition. It turns out that in this model, the speed of light has to be added to the local expansion speed of the universe, this is just a result of the particular choice of coordinates.

    I thought I had understood everything pretty well. The two models explain things differently, but consistently. For example, in the first model, time at very distant objects grinds to a halt as they approach (but never exceed) the speed of light. They will never reach our age. In the second model, they are the same age as us, but light from those events will never reach us because space between us is expanding faster than the speed of light. Every ray of light that tries to travel towards us, is actually retreating away from us because the distance in between increases so rapidly. So in both cases, we will never see those events.

    However, I then tried to think what would happen if the universe would come back together in a big crunch. I know that it has recently been discovered that this will probably never happen, since the expansion is accellerating, but with enough mass and no dark energy this could be theoretically possible (so I'm told), and should not lead to a contradiction in the models.

    The problem is that the first model results in a twin paradox, without the asymmetry that normally solves the paradox: any observer can consider himself to be in the center, and will be convinced that time at other galaxies is running more slowly.

    Obviously the Cosmological Model does not suffer from this paradox because by definition it has no time dilation. But the first model should be a valid choice of coordinates as well, so how is the paradox solved there?

    I thought for a moment that, if you consider our position to be in the center of the universe, it would also be in the center of the cosmological field of gravity that is pulling everything back together, and the gravitational time dilation might exactly offset the time dilation caused by speed. Time is moving more slowly for them because they are moving, but then again faster because they are higher "up" in the field of gravity. The two effects are not constant (speed and "height" change all the time, so sometimes distant time would be running faster (e.g. at the moment of reversal), and other times slower), but at the big crunch all the clocks should be back in sync. Has anyone thought of this before and solved it, or am I on a dead end track?

    In the other thread I asked about clocks being tossed up hoping that the answer would be "they show the same time" and that would solve the problem. Only that seemed to be an oversimplification, since the metric of space does not resemble that of a planet (Schwarzschild, FLRW,...).

    So, any ideas? I'm sure someone has thought of all this before and solved it...
     
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  3. Jul 28, 2009 #2

    Ich

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    Interesting question. From my understanding, this treatment must give the correct null result, but I haven't worked the problem yet.
     
  4. Jul 28, 2009 #3

    Dale

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    To the best of my understanding you cannot use a flat metric to describe the expanding universe. I think that the FLRW metric is required. So, 1 is not applicable. I don't know about your description of the FLRW metric in 2.
     
  5. Jul 29, 2009 #4

    Fredrik

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    It's actually not correct that we can't see objects that move away from us faster than c. There are several threads about that in the cosmology forum.

    In the center of what? The universe? It doesn't have a center.

    I'm not sure I understand your first "model".

    Apparently you did mean center of the universe. It's like talking about the center of the surface of a sphere. Quite literally actually, because in the FLRW solutions with a big crunch, "space at time t" is a 3-sphere.

    I don't understand why you think there appears to be a twin paradox in the big crunch solutions.
     
  6. Jul 29, 2009 #5

    Ich

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    Fredrik, this problem is exactly what we're discussing in the other thread. Of course the univers doesn't have a center, but you can start building a coordinate system with your position at the center. That system is useful to second order in distance if we introduce the concepts of SR and gravitational time dilation. But bot must cancel in the long run. If not, either the description is invalid (contrary to my claims), or there is a paradox.
     
  7. Jul 29, 2009 #6
    If I understand you correctly, an analogous situation would be just two equal mass planets in deep space initially moving away from each other at a slowing pace due to gravity, each with a clock.

    If that's what you mean, if the clocks were synchronized initially while the planets were close, they should read the same when they meet again due to gravity, if we consider the clocks approximately "local" at the start and stop events. There would be no asymmetry and no difference in elapsed time between the clocks.
     
  8. Jul 29, 2009 #7

    Ich

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    I think I solved it, in second order approximation. The contributions [tex]\frac{d \Delta t}{d\phi}[/tex] seem to be constant A/2 for gravitation, and [tex]A \,cos(\phi)^2[/tex] for the velocity contribution (with some arbitrary A, and phi the parameter of the parametric overdense FRW solution w/o Lambda).
    I calculated sloppily, and stopped thinking immediately after my first try gave the result I expected. Maybe someone could check.
     
  9. Jul 29, 2009 #8

    Dale

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    Hi michelcoleman,

    Let's use a reasonably close* geometric analogy, let's consider the geometry on the surface of a sphere. We will consider the south pole to be the big bang and the north pole to be the big crunch. Lattitude lines are lines of constant coordinate time, and longitude lines are lines of constant coordinate position, so this represents a universe with a single spatial dimension. Note, that all of the longitude lines are geodesics (great circles) so they can represent observers who travel inertially from the bang to the crunch. Note also, that because of the symmetry all of the longitude lines have the same length. So any pair of "bang-crunch" inertial twins will record the same time, by symmetry, despite the fact that they are all always "accelerating" wrt each other. They all agree on this due to the symmetry of the metric across the universe.

    Now, consider a different geometry, that of the surface of a hemisphere cut through both poles, one side of the universe is now curved and the other side is flat. Now, the diameter of the flat side going from pole to pole is also a "bang-crunch" geodesic, but this geodesic is shorter than any of the longitude geodesics. Here, again, all "bang-crunch" twins are inertial, but now the universe itself is asymmetric. All twins agree on the asymmetry of the universe's metric so they all agree that the flat geodesic is shorter than the longitude geodesics.

    So for the twins to record different times there must be some asymmetry that is agreed upon by all twins. Either some asymmetry in their proper acceleration, or some asymmetry in the metric itself.

    *The analogy is not perfect because in a close neighborhood of any point the metric is Euclid's rather than Minkowski's. But it should allow a reasonable grasp of the geometric concepts involved.
     
  10. Jul 30, 2009 #9
    Yes, I do agree it makes sense that the clocks in the end must match, because of the obvious symmetry of the situation. The problem is, since the other galaxies are moving, their time must be going more slowly, so something has to compensate for that in order for the indicated times to be equal again in the end. There was never any question over whether or not the times would be equal. Just how this comes about mathematically considering time dilation from any particular point of view.

    But it looks like Ich has cracked it: The variable time dilation due to speed, and the constant time dilation due to gravity (apparently), when integrated from bang to crunch, exactly cancel each other out!

    So right after the big bang, time at a distance is running more slowly, then while the expansion slows down, time at a distance starts moving more rapidly, gravitational time dilation becoming more dominant (our time is slower because we are in the middle of the field from our point of view), and the clocks are in sync again when the expansion turns into collapse. During the initial part of the collapse, distant clocks actually get ahead of ours, but finally speed kicks in again so the distant clocks become slower than ours again (slower moving, but still ahead) and exactly match our time at the crunch!

    If Ich's calculations are correct, this does indeed solve the problem.
     
  11. Aug 3, 2009 #10

    Ich

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    I checked my calculations again. They seem to be correct, with the factor
    [tex]A=\frac{H_0 \Omega}{2\sqrt{\Omega-1}}[/tex]
    One correction: it's not [tex] A \,cos(\phi)^2 [/tex], I meant [tex] A \,cos(\phi/2)^2 [/tex] instead.
     
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